HW#8 - 1-1 neither yes 11.4 Mon.subseq SL lim sup lim inf...

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SOLUTIONS TO HOMEWORK # 8 11.1 (a) 1, 5, 1, 5, 1, 5, 1, 5. (b) The sequence 5 , 5 , 5 ,... is constant. The selection function is σ ( k ) = 2 k . 11.2 Monot.subseq. SL lim sup, lim inf Conv/Diverg bounded a n a 2 k = 1 , 1 ,... {- 1 , 1 } 1,-1 neither yes b n b n itself { 0 } 0,0 conv. yes c n c n itself {∞} , div.to no d n d n itself { 6 7 } 6 7 , 6 7 conv. yes 11.3 Mon.subseq. SL lim sup, lim inf Co. .. bounded s n s 6 k = 1 , 1 ,... { 1 , 1 2 , - 1 2 , - 1 } 1,-1 neither yes t n t n itself { 0 } 0,0 conv. yes u n u 2 n = 1 4 , 1 16 ,... { 0 } 0 , 0 conv. yes v n v 2 n = 3 2 , 5 4 ,... {- 1 , 1 }
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Unformatted text preview: 1 ,-1 neither yes 11.4 Mon.subseq. SL lim sup, lim inf Co. .. bounded w n w 2 k = 4 , 16 ,... {-∞ , ∞} ∞ ,-∞ neither no x n x 2 k = 5 , 5 ,... { 1 5 , 5 } 5 , 1 5 neitther yes y n y 2 k = 2 , 2 ,... { , 2 } 2 , neither yes z n z 8 k = 8 , 16 ,... {-∞ , , ∞} ∞ ,-∞ neither no 1...
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This note was uploaded on 04/28/2008 for the course MATH 312 taught by Professor Olsavskygregorm during the Spring '08 term at Penn State.

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